# The Stacks Project

## Tag 023N

Effective descent for modules along faithfully flat ring maps.

Proposition 34.3.9. Let $R \to A$ be a faithfully flat ring map. Then

1. any descent datum on modules with respect to $R \to A$ is effective,
2. the functor $M \mapsto (A \otimes_R M, can)$ from $R$-modules to the category of descent data is an equivalence, and
3. the inverse functor is given by $(N, \varphi) \mapsto H^0(s(N_\bullet))$.

Proof. We only prove (1) and omit the proofs of (2) and (3). As $R \to A$ is faithfully flat, there exists a faithfully flat base change $R \to R'$ such that $R' \to A' = R' \otimes_R A$ has a section (namely take $R' = A$ as in the proof of Lemma 34.3.6). Hence, using Lemma 34.3.8 we may assume that $R \to A$ has a section, say $\sigma : A \to R$. Let $(N, \varphi)$ be a descent datum relative to $R \to A$. Set $$M = H^0(s(N_\bullet)) = \{n \in N \mid 1 \otimes n = \varphi(n \otimes 1)\} \subset N$$ By Lemma 34.3.7 it suffices to show that $A \otimes_R M \to N$ is an isomorphism.

Take an element $n \in N$. Write $\varphi(n \otimes 1) = \sum a_i \otimes x_i$ for certain $a_i \in A$ and $x_i \in N$. By Lemma 34.3.2 we have $n = \sum a_i x_i$ in $N$ (because $\sigma^0_0 \circ \delta^1_1 = \text{id}$ in any cosimplicial object). Next, write $\varphi(x_i \otimes 1) = \sum a_{ij} \otimes y_j$ for certain $a_{ij} \in A$ and $y_j \in N$. The cocycle condition means that $$\sum a_i \otimes a_{ij} \otimes y_j = \sum a_i \otimes 1 \otimes x_i$$ in $A \otimes_R A \otimes_R N$. We conclude two things from this. First, by applying $\sigma$ to the first $A$ we conclude that $\sum \sigma(a_i) \varphi(x_i \otimes 1) = \sum \sigma(a_i) \otimes x_i$ which means that $\sum \sigma(a_i) x_i \in M$. Next, by applying $\sigma$ to the middle $A$ and multiplying out we conclude that $\sum_i a_i (\sum_j \sigma(a_{ij}) y_j) = \sum a_i x_i = n$. Hence by the first conclusion we see that $A \otimes_R M \to N$ is surjective. Finally, suppose that $m_i \in M$ and $\sum a_i m_i = 0$. Then we see by applying $\varphi$ to $\sum a_im_i \otimes 1$ that $\sum a_i \otimes m_i = 0$. In other words $A \otimes_R M \to N$ is injective and we win. $\square$

The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 641–655 (see updates for more information).

\begin{proposition}
\label{proposition-descent-module}
\begin{slogan}
Effective descent for modules along faithfully flat ring maps.
\end{slogan}
Let $R \to A$ be a faithfully flat ring map.
Then
\begin{enumerate}
\item any descent datum on modules with respect to $R \to A$
is effective,
\item the functor $M \mapsto (A \otimes_R M, can)$ from $R$-modules
to the category of descent data is an equivalence, and
\item the inverse functor is given by $(N, \varphi) \mapsto H^0(s(N_\bullet))$.
\end{enumerate}
\end{proposition}

\begin{proof}
We only prove (1) and omit the proofs of (2) and (3).
As $R \to A$ is faithfully flat, there exists a faithfully flat
base change $R \to R'$ such that $R' \to A' = R' \otimes_R A$ has
a section (namely take $R' = A$ as in the proof of
Lemma \ref{lemma-ff-exact}). Hence, using
Lemma \ref{lemma-descent-descends}
we may assume that $R \to A$ has a section, say $\sigma : A \to R$.
Let $(N, \varphi)$ be a descent datum relative to $R \to A$.
Set
$$M = H^0(s(N_\bullet)) = \{n \in N \mid 1 \otimes n = \varphi(n \otimes 1)\} \subset N$$
By Lemma \ref{lemma-recognize-effective} it suffices to show that
$A \otimes_R M \to N$ is an isomorphism.

\medskip\noindent
Take an element $n \in N$. Write
$\varphi(n \otimes 1) = \sum a_i \otimes x_i$ for certain
$a_i \in A$ and $x_i \in N$. By Lemma \ref{lemma-descent-datum-cosimplicial}
we have $n = \sum a_i x_i$ in $N$ (because
$\sigma^0_0 \circ \delta^1_1 = \text{id}$ in any cosimplicial object).
Next, write $\varphi(x_i \otimes 1) = \sum a_{ij} \otimes y_j$ for
certain $a_{ij} \in A$ and $y_j \in N$.
The cocycle condition means that
$$\sum a_i \otimes a_{ij} \otimes y_j = \sum a_i \otimes 1 \otimes x_i$$
in $A \otimes_R A \otimes_R N$. We conclude two things from this.
First, by applying $\sigma$ to the first $A$ we conclude that
$\sum \sigma(a_i) \varphi(x_i \otimes 1) = \sum \sigma(a_i) \otimes x_i$
which means that $\sum \sigma(a_i) x_i \in M$. Next, by applying
$\sigma$ to the middle $A$ and multiplying out we conclude that
$\sum_i a_i (\sum_j \sigma(a_{ij}) y_j) = \sum a_i x_i = n$. Hence
by the first conclusion we see that $A \otimes_R M \to N$ is
surjective. Finally, suppose that $m_i \in M$ and
$\sum a_i m_i = 0$. Then we see by applying $\varphi$ to
$\sum a_im_i \otimes 1$ that $\sum a_i \otimes m_i = 0$.
In other words $A \otimes_R M \to N$ is injective and we win.
\end{proof}

Comment #874 by Bhargav Bhatt on August 3, 2014 a 6:46 pm UTC

Suggested slogan: One has effective descent for modules along faithfully flat ring maps.

Comment #2832 by Ko Aoki on October 2, 2017 a 6:57 am UTC

Two typos in the proof: "we may assume that $R \to A$ as a section" should be replaced by "we may assume that $R \to A$ has a section", and "(because $\sigma^0_0 \circ \delta^1_0 = \text{id}$ in any cosimplicial object)" should be replaced by "(because $\sigma^0_0 \circ \delta^1_1 = \text{id}$ in any cosimplicial object)"

Comment #2931 by Johan (site) on October 10, 2017 a 1:57 am UTC

THanks, fixed here.

There are also 2 comments on Section 34.3: Descent.

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